Consider tan(2x) = \(\displaystyle \sqrt3\) for 0 \(\displaystyle \leq x < 2\pi\)

reference angle x = \(\displaystyle \tan^{-1}\)(\(\displaystyle \sqrt{3}\)) = \(\displaystyle \frac{\pi}{3}\)

In the given domain 0 \(\displaystyle \leq x < 2\pi\), the positive ratio of tangent is in quadrants I and III. So the two standard position angles in this interval are

2x = \(\displaystyle \frac{\pi}{3}\) and \(\displaystyle \frac{4\pi}{3}\)

Now, divide both sides by 2. Then I get

x = \(\displaystyle \frac{\pi}{6}\) and \(\displaystyle \frac{2\pi}{3}\) --------- (*)

The period of tangent is: period = \(\displaystyle \frac{\pi}{b}\)

where b = 2 in tan(2x).

So the period is \(\displaystyle \frac{\pi}{2}\).

Then, add this period to each solution of x in the first cycle to get the next two solutions in the second cycle. So I get

x = \(\displaystyle \frac{2\pi}{3}\) and \(\displaystyle \frac{7\pi}{6}\)

Therefore, the solutions are x=\(\displaystyle \frac{\pi}{6}\), \(\displaystyle \frac{2\pi}{3}\), \(\displaystyle \frac{2\pi}{3}\) and \(\displaystyle \frac{7\pi}{6}\)

The answer key answer says x = \(\displaystyle \frac{\pi}{6}\), \(\displaystyle \frac{2\pi}{3}\), \(\displaystyle \frac{7\pi}{6}\) and \(\displaystyle \frac{5\pi}{3}\).

I am very confused. Can someone explain this? Thanks a lot.

reference angle x = \(\displaystyle \tan^{-1}\)(\(\displaystyle \sqrt{3}\)) = \(\displaystyle \frac{\pi}{3}\)

In the given domain 0 \(\displaystyle \leq x < 2\pi\), the positive ratio of tangent is in quadrants I and III. So the two standard position angles in this interval are

2x = \(\displaystyle \frac{\pi}{3}\) and \(\displaystyle \frac{4\pi}{3}\)

Now, divide both sides by 2. Then I get

x = \(\displaystyle \frac{\pi}{6}\) and \(\displaystyle \frac{2\pi}{3}\) --------- (*)

The period of tangent is: period = \(\displaystyle \frac{\pi}{b}\)

where b = 2 in tan(2x).

So the period is \(\displaystyle \frac{\pi}{2}\).

Then, add this period to each solution of x in the first cycle to get the next two solutions in the second cycle. So I get

x = \(\displaystyle \frac{2\pi}{3}\) and \(\displaystyle \frac{7\pi}{6}\)

Therefore, the solutions are x=\(\displaystyle \frac{\pi}{6}\), \(\displaystyle \frac{2\pi}{3}\), \(\displaystyle \frac{2\pi}{3}\) and \(\displaystyle \frac{7\pi}{6}\)

**The two middle solutions are the same.**The answer key answer says x = \(\displaystyle \frac{\pi}{6}\), \(\displaystyle \frac{2\pi}{3}\), \(\displaystyle \frac{7\pi}{6}\) and \(\displaystyle \frac{5\pi}{3}\).

**I actually found my mistake.**If I added \(\displaystyle \pi\) instead of \(\displaystyle \frac{\pi}{2}\) in (*) above, I would get the right solutions in the second cycle. But, I don't understand why you have to do that, since the period of tan(2X) is \(\displaystyle \frac{\pi}{b}\) = \(\displaystyle \frac{\pi}{2}\).I am very confused. Can someone explain this? Thanks a lot.

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